Click on the arrow to see answer.
Suppose \(B\) is also an additive inverse of \(A\). Then
\[ -A=-A+\left ( A+B\right ) =\left ( -A+A\right ) +B=0+B=B \]
Solved Problems for Chapter 4
Using only the properties given in Theorem ?? and Theorem ??, show that the
additive inverse of \(A\), \(-A\), is unique.
Using only the properties given in Theorem ?? and Theorem ??, show that the \(n\times m\)
zero matrix, \(O\), is unique.
Click on the arrow to see answer.
Suppose \(O^{\prime }\) is also an \(n\times m\) additive identity. Then \(O^{\prime }=O^{\prime }+O=O.\)
Using only the properties given in Theorem ?? and Theorem ??, show that
\(0A=O.\)
Click on the arrow to see answer.
\(0A=\left ( 0+0\right ) A=0A+0A.\) Now add \(-\left ( 0A\right ) \) to both sides. Then \(O=0A\).
Using only the properties given in Theorem ?? and Theorem ??, as well as
previous problems, show \(\left ( -1\right ) A=-A.\)
Click on the arrow to see answer.
\(A+\left ( -1\right ) A=\left ( 1+\left ( -1\right ) \right ) A=0A=O.\) Therefore, from the uniqueness of the additive
inverse proved in the above Problem ??, it follows that \( -A=\left ( -1\right ) A\).
Consider the matrices \( A =\left [ \begin{array}{rrr} 1 & 2 & 3 \\ 2 & 1 & 7 \end{array} \right ], B=\left [ \begin{array}{rrr} 3 & -1 & 2 \\ -3 & 2 & 1 \end{array} \right ], C =\left [ \begin{array}{rr} 1 & 2 \\ 3 & 1 \end{array} \right ], \\ D=\left [ \begin{array}{rr} -1 & 2 \\ 2 & -3 \end{array} \right ], E=\left [ \begin{array}{r} 2 \\ 3 \end{array} \right ]\).
Find the following if possible. If it is not possible explain why.
- (a)
- \(-3A\)
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\(\left [ \begin{array}{rrr} -3 & -6 & -9 \\ -6 & -3 & -21 \end{array} \right ]\) - (b)
- \(3B-A\)
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\(\left [ \begin{array}{rrr} 8 & -5 & 3 \\ -11 & 5 & -4 \end{array} \right ]\) - (c)
- \(AC\)
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Not possible - (d)
- \(CB\)
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\(\left [ \begin{array}{rrr} -3 & 3 & 4 \\ 6 & -1 & 7 \end{array} \right ]\) - (e)
- \(AE\)
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Not possible - (f)
- \(EA\)
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Not possible
Let \(A=\left [ \begin{array}{rr} -1 & -1 \\ 3 & 3 \end{array} \right ] \). Find all \(2\times 2\) matrices, \(B\) such that \(AB=O.\)
Click on the arrow to see answer.
\[ \left [ \begin{array}{rr} -1 & -1 \\ 3 & 3 \end{array} \right ] \left [ \begin{array}{cc} x & y \\ z & w \end{array} \right ] =\left [ \begin{array}{cc} -x-z & -w-y \\ 3x+3z & 3w+3y \end{array} \right ] =\left [ \begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array} \right ] \]
Solution is: \( w=-y,x=-z \) so the matrices are of the form \(\left [ \begin{array}{rr} x & y \\ -x & -y \end{array} \right ].\) Let \(A=\left [ \begin{array}{rr} 1 & 2 \\ 3 & 4 \end{array} \right ] ,B=\left [ \begin{array}{rr} 1 & 2 \\ 3 & k \end{array} \right ] .\) Is it possible to choose \(k\) such that \(AB=BA?\) If so, what should \(k\) equal?
Click on the arrow to see answer.
\[ \left [ \begin{array}{cc} 1 & 2 \\ 3 & 4 \end{array} \right ] \left [ \begin{array}{cc} 1 & 2 \\ 3 & k \end{array} \right ] = \left [ \begin{array}{cc} 7 & 2k+2 \\ 15 & 4k+6 \end{array} \right ] \]
\[ \left [ \begin{array}{cc} 1 & 2 \\ 3 & k \end{array} \right ] \left [ \begin{array}{cc} 1 & 2 \\ 3 & 4 \end{array} \right ] = \left [ \begin{array}{cc} 7 & 10 \\ 3k+3 & 4k+6 \end{array} \right ] \]
Thus you must have \( \begin{array}{c} 3k+3=15 \\ 2k+2=10 \end{array} \). Therefore \(k=4\). Find \(2 \times 2\) matrices \(A\) and \(B\) such that \(A \neq O\) and \(B \neq O\) but \(AB = O\).
Click on the arrow to see answer.
Let \(A = \left [ \begin{array}{rr} 1 & -1 \\ -1 & 1 \end{array} \right ], B = \left [ \begin{array}{cc} 1 & 1 \\ 1 & 1 \end{array} \right ].\)
\[ \left [ \begin{array}{rr} 1 & -1 \\ -1 & 1 \end{array} \right ] \left [ \begin{array}{cc} 1 & 1 \\ 1 & 1 \end{array} \right ] = \left [ \begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array} \right ] \]
Suppose \(AB=AC\) and \(A\) is an invertible \(n\times n\) matrix. Does it follow that \(B=C?\) Explain why or why
not.
Click on the arrow to see answer.
Yes \(B=C\). Multiply \(AB = AC\) on the left by \(A^{-1}\).
Let
\begin{equation*} A=\left [ \begin{array}{rrr} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 0 & 2 \end{array} \right ] \end{equation*}
Find \(A^{-1}\) if possible. If \(A^{-1}\) does not exist, explain why.
Click on the arrow to see answer.
\(\left [ \begin{array}{ccc} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 0 & 2 \end{array} \right ]^{-1}= \left [ \begin{array}{rrr} -2 & 4 & -5 \\ 0 & 1 & -2 \\ 1 & -2 & 3 \end{array} \right ]\)
Let
\begin{equation*} A=\left [ \begin{array}{rrr} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 4 & 5 & 10 \end{array} \right ] \end{equation*}
Find \(A^{-1}\) if possible. If \(A^{-1}\) does not exist, explain why.
Click on the arrow to see answer.
The reduced row echelon form is \(\left [ \begin{array}{ccc} 1 & 0 & \frac {5}{3} \\ 0 & 1 & \frac {2}{3} \\ 0 & 0 & 0 \end{array} \right ]\). There is no
inverse.
Show that if \(A^{-1}\) exists for an \(n\times n\) matrix, then it is unique. That is, if \(BA=I\) and \(AB=I,\) then
\(B=A^{-1}.\)
Click on the arrow to see answer.
\(A^{-1}=A^{-1}I=A^{-1}\left ( AB\right ) =\left ( A^{-1}A\right ) B=IB=B.\)
Show that if \(A\) is an invertible \(n\times n\) matrix, then so is \(A^{T} \) and \(\left ( A^{T}\right ) ^{-1}=\left ( A^{-1}\right ) ^{T}.\)
Click the arrow to see answer.
You need to show that \(\left ( A^{-1}\right ) ^{T}\) acts like the inverse of \(A^{T} \) because from uniqueness in
the above problem, this will imply it is the inverse. From properties of the
transpose,
\[ A^{T}\left ( A^{-1}\right ) ^{T} =\left ( A^{-1}A\right ) ^{T}=I^{T}=I \]
\[\left ( A^{-1}\right ) ^{T}A^{T} =\left ( AA^{-1}\right ) ^{T}=I^{T}=I \]
Hence \(\left ( A^{-1}\right ) ^{T}=\left ( A^{T}\right ) ^{-1}\) and this last matrix exists. Show \(\left ( AB\right ) ^{-1}=B^{-1}A^{-1}\) by verifying that
\begin{equation*} AB\left ( B^{-1}A^{-1}\right ) =I \end{equation*}
and \begin{equation*} B^{-1}A^{-1}\left ( AB\right ) =I \end{equation*}
Click on the arrow to see answer.
\(\left ( AB\right ) B^{-1}A^{-1}=A\left ( BB^{-1}\right ) A^{-1}=AA^{-1}=I\) \(B^{-1}A^{-1}\left ( AB\right ) =B^{-1}\left ( A^{-1}A\right ) B=B^{-1}IB=B^{-1}B=I\)
Find an \(LU\) factorization of the coefficient matrix and use it to solve the system of
equations.
\begin{equation*} \begin{array}{c} x+2y=5 \\ 2x+3y=6 \end{array} \end{equation*}
Click on the arrow to see answer.
An \(LU\) factorization of the coefficient matrix is
\[ \left [ \begin{array}{cc} 1 & 2 \\ 2 & 3 \end{array} \right ] = \left [ \begin{array}{cc} 1 & 0 \\ 2 & 1 \end{array} \right ] \left [ \begin{array}{cc} 1 & 2 \\ 0 & -1 \end{array} \right ] \]
First
solve \[ \left [ \begin{array}{cc} 1 & 0 \\ 2 & 1 \end{array} \right ] \left [ \begin{array}{c} u \\ v \end{array} \right ] =\left [ \begin{array}{c} 5 \\ 6 \end{array} \right ] \]
which gives \(\left [ \begin{array}{c} u \\ v \end{array} \right ] =\) \(\left [ \begin{array}{r} 5 \\ -4 \end{array} \right ] .\) Then solve \[ \left [ \begin{array}{rr} 1 & 2 \\ 0 & -1 \end{array} \right ] \left [ \begin{array}{c} x \\ y \end{array} \right ] =\left [ \begin{array}{r} 5 \\ -4 \end{array} \right ] \]
which says that \(y=4\) and \(x=-3.\) Find an \(LU\) factorization of the coefficient matrix and use it to solve the system of
equations.
\begin{equation*} \begin{array}{c} x+2y+z=1 \\ y+3z=2 \\ 2x+3y=6 \end{array} \end{equation*}
Click on the arrow to see answer.
An \(LU\) factorization of the coefficient matrix is
\[ \left [ \begin{array}{rrr} 1 & 2 & 1 \\ 0 & 1 & 3 \\ 2 & 3 & 0 \end{array} \right ] = \left [ \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 2 & -1 & 1 \end{array} \right ] \left [ \begin{array}{rrr} 1 & 2 & 1 \\ 0 & 1 & 3 \\ 0 & 0 & 1 \end{array} \right ] \]
First
solve \[ \left [ \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 2 & -1 & 1 \end{array} \right ] \left [ \begin{array}{c} u \\ v \\ w \end{array} \right ] =\left [ \begin{array}{c} 1 \\ 2 \\ 6 \end{array} \right ] \]
which yields \(u=1,v=2,w=6\). Next solve \[ \left [ \begin{array}{rrr} 1 & 2 & 1 \\ 0 & 1 & 3 \\ 0 & 0 & 1 \end{array} \right ] \left [ \begin{array}{c} x \\ y \\ z \end{array} \right ] =\left [ \begin{array}{c} 1 \\ 2 \\ 6 \end{array} \right ] \]
This yields \(z=6,y=-16,x=27.\) Bibliography
These problems come from the end of Chapter 2 of Ken Kuttler’s A First Course in Linear Algebra. (CC-BY)
Ken Kuttler, A First Course in Linear Algebra, Lyryx 2017, Open Edition, pp. 90–98, 104–106.